Numbers
Helmer Aslaksen
Dept. of Teacher Education & Dept. of Mathematics
University of Oslo
[email protected]
www.math.nus.edu.sg/aslaksen/
Natural numbers
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Let N denote the positive, natural numbers {1, 2, 3, . . .}.
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Remember that positive means > 0 and negative means < 0,
so nonnegative is not the same as positive, but means positive
or zero.
Why is (−1)(−1) = 1?
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One way to understand why (−1)(−1) = 1, is to say that
multiplying by −1 is the same as “flipping” across zero on the
number line, in which case, flipping twice does nothing.
However, it is instructive to also see an algebraic proof. Assume
that we know how to multiply natural numbers, and that we want
to extend this to integers. We want to do this in such a way that
the following three properties are preserved.
1. Commutative ab = ba
2. Associative (ab)c = a(bc )
3. Distributive a(b + c ) = ab + ac
Why is (−1)(−1) = 1? 2
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Assume that a, b ∈ N. We know that
a(−b) = (−b) + (−b) + · · · + (−b) = −ab
(1)
by repeated addition.
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To compute (−a)b, we use commutativity and Equation (1) to
get
(−a)b = b(−a) = −ba = −ab.
Why is (−1)(−1) = 1? 3
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We want to show that (−1)(−1) = 1, and to do that, we
consider (−1)(−1) − 1 and use distributivity
(−1)(−1) − 1 =
(−1)(−1) + (−1) =
(−1)(−1) + (−1) · 1 =
(−1)(−1 + 1) =
(−1) · 0 = 0.
Hence (−1)(−1) = 1.
Division by zero
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The key to understanding division and fractions is that
a
= c ⇐⇒ a = bc .
b
(2)
This shows why we cannot divide by 0. If b = 0, we get
a
= c ⇐⇒ a = 0 · c = 0,
0
which shows that we get a contradiction if we try to assign a
value to a/0 when a 6= 0.
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But what if a = 0? In that case, the above equation just says
that 0 = 0 · c = 0, which is true for any c. But that is precisely
the problem. We could theoretically define 0/0 to be anything,
without violating (2), but which value should we choose? Since
we theoretically could pick any value, we say that 0/0 is an
indeterminate form.
Division by fractions
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Many students do not understand why dividing by a fraction is
the same as inverting the second fraction and multiplying
a c
ad
: =
.
b d
bc
(3)
To see this, we must show that if multiply the number on the
right by the divisor, we get the dividend, i.e.,
ad
bc
c
a
=
d
b
dc
cd
=
a
.
b
Division by fractions 2
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Another way to see this is to use complex fractions
a
bd
ad
b
=
.
c
bc
bd
d
Division by fractions 3
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It is also instructive to consider unit fractions, 1/d. There are
many ways to argue that
a:
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1
= ad .
d
It then follows from associativity that
a c
1
1
ad
1
: =
a :
: c = (ad ) : c =
.
b
d
b
d
b
bc
Powers
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Assume that we have defined an with n ∈ N to be
n
n
z }| {
a = a · ... · a.
(4)
For n, m ∈ N it is easy to see that we have the following
property
n
m
n+m
z }| { z }| { z }| {
an am = a · · · a a · · · a = a · · · a = an+m .
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(5)
We now want to extend Definition (4) to n ∈ Z in way that
preserves property (5). In other words, we will assume that (5)
holds, and see what that implies about a0 and a−n for n ∈ N.
Powers 2
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Setting m = 0 in (5), we get
an = an+0 = an · a0 ,
so if a 6= 0, we can divide by an and conclude that a0 = 1. (We
will discuss 00 later.)
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We now set m = −n in (5) and get
1 = a 0 = a n −n = a n a −n ,
so it follows that
a −n =
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1
.
an
We can now check that
an
an−m am
=
= an−m
m
m
a
a
holds for n, m ∈ Z.
Powers 3
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Another way to understand this, is to interpret a0 as an “empty
product”. The “empty sum” 0 · a is the additive identity 0, while
the “empty product” a0 is the multiplicative identity 1.
Fractional Exponents
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Again we want to extend a definition to a larger set of numbers
by preserving a property. We know that for m, n ∈ N we have
n
n ·m
n
z }| {
z }| {
z }| {
(an )m = |an ·{z
· · a}n = (a · · · a) · · · (a · · · a) = a · · · a = an·m . (6)
|
{z
}
m
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m
We want to extend the definition of an to n ∈ Q, while
maintaining property (6). To see how to do this, we simply write
x = a1/n .
Then
1
x n = (a1/n )n = a( n n) = a1 = a
so we see that
a1/n =
I
√
n
a.
Using property (6) again, we get that
1
am/n = (am ) n =
√
n
am .
Is 00 = 1?
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We have seen that for a 6= 0 we have a0 = 1, so lima→0 a0 = 1.
It therefore seems natural to define 00 = 1. However, for x > 0,
we have 0x = 0 and it follows that limx →0+ 0x = 0.
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This shows that the function f (a, x ) = ax does not have a limit
at (0, 0) since we get different values depending on how we
approach (0, 0). It follows that f is not continuous at (0, 0).
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That makes it harder to find a good value for 00 , but not
impossible.
Is 00 = 1? 2
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We often write a polynomial as
p(x ) =
n
X
ak x k .
k =0
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However, then
p(0) = a0 00 + · · · + an 0n = a0 00 ,
and we are implicitly assuming that 00 = 1.
Is 00 = 1? 3
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We can also consider a power series like
f (x ) =
1
1−x
Then
f (0) = 1 =
=
∞
X
∞
X
x n.
n =0
0n = 00 .
n=0
If we do not define 00 to be 1, we will have trouble with even
simple expressions like this.
Is 00 = 1? 4
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Another example is the Binomial Theorem
(a + b)n =
n X
n
k =0
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k
a k b n −k .
Setting a = 0 on both sides and assuming b 6= 0 we get
bn = (0 + b)n =
n X
n
k =0
k
0k bn−k =
n 0 n
0 b = 00 bn ,
0
where, we have used that 0k = 0 for k > 0, and that
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n
0
= 1.
We see that we must set 00 = 1 in order for the binomial
theorem to be valid.
Is 00 = 1? 5
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Another reason why 00 = 1 is because x 0 is the “empty
product”, which should be the multiplicative identigy 1. For the
same reason we also get 0! = 1.
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In order for the differentiation rule
d n
x = nx n−1
dx
to hold for n = 1 when x = 0, we get
1=
d
d 1
x=
x = 1 · x 1− 1 = x 0 ,
dx
dx
which requires 00 = 1.
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So to sum up, we must write 00 = 1 to make many expressions
work.
Rational numbers
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We will study the rational numbers
o
na
Q=
a, b ∈ Z, b 6= 0 .
b
We want to show that
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√
2 is irrational.
We will need the following lemma
Lemma
A natural number a is even if and only if a2 is even.
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Proof: If a is even we can write a = 2k with k ∈ Z and then
a2 = (2k )2 = 4k 2 = 2(2k 2 ),
so we see that
a is even =⇒ a2 is even.
Rational numbers 2
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In order to show
a is even ⇐= a2 is even,
we will use that
p =⇒ q
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is the same as ¬p ⇐= ¬q .
So we will show that
a is odd =⇒ a2 is odd.
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If a = 2k + 1 with k ∈ Z, then
a2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k ) + 1,
so a2 is odd.
Rational numbers 3
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Theorem
√
2 is irrational.
√
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Proof: We will assume that 2 is rational and can be written as
a/b, where a, b ∈ Z are relatively prime, i.e., they have no
common factors. Then
a2
2= 2
b
and
2b2 = a2 ,
and we see that a2 is even. But then we know from the above
Lemma that a is also even, so a = 2k with k ∈ Z and
a2 = (2k )2 = 4k 2 = 2(b2 )
or
b2 = 2k 2 .
Since b2 is even, it follows that b is also even. We have now
shown that both a and b are even, but this contradicts the
assumption that a and b are relatively prime.
Countable
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We say that two sets X and Y have the same cardinality if there
is a bijection f : X → Y .
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We say that X is countably infinite if there is a bijection
f : N → X.
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We say that X is countable if there is a surjection f : N → X .
This means that we can write the elements of X as a list.
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A countable set is either countably infinite or finite.
Countable 2
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The set of integers, Z is countable, since
Z = {0, 1, −1, 2, −2, 3, −3, ...}.
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The set of rational numbers, Q, is countable. This can be seen
in many ways.
Countable 3
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The rational numbers a/b correspond to the pair (a, b), so Q
corresponds to Z × (Z − {0}).
Countable 4
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This shows that the set of positive rationals is countable. By
alternating between positive and negative numbers, we can
show that the whole of Q is countable.
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This picture also shows that a countable union of countable sets
is countable.
Countable 5
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In 1874, Georg Cantor (1845 -– 1918) proved that R is not
countable.
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Assume that R is countable. Then [0, 1] is also countable, and
we can write [0, 1] = {r1 , r2 , . . .} where ri = 0.di1 di2 . . ..
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We then construct a number r = 0.d1 d2 . . ., where di 6= dii and
di 6= 9. Then r 6= ri for all i, and r is not in the list.